Subject: Chemistry — Chemical Thermodynamics
Question: 2 mole of $Hg(g)$ is combusted in a fixed volume bomb
calorimeter with excess of $O_{2}$ at $298\ K$ and 1 atm into $HgO(s)$.
During the reaction, temperature increases from $298.0\ K$ to
$312.8\ K$. If heat capacity of the bomb calorimeter and enthalpy of
formation of $Hg(g)$ are $20.00\ kJ\ K^{- 1}$ and $61.32\ kJ\ mol^{- 1}$
at $298\ K$, respectively, the calculated standard molar enthalpy of
formation of $HgO(s)$ at $298\ K$ is $X\ kJ\ mol^{- 1}$. The value of
$|X|$ is \_\_\_\_\_\_\_\_\_\_.
\[Given: Gas constant $R = 8.3\ JK^{- 1}\ mol^{- 1}$\]
Answer: 90.39
Solution: [IMAGE] $$q_{combustion} = \left( C_{T} \right)\mathrm{\Delta}T$$
$$= 20 \times \lbrack 312.8 - 298\rbrack$$
$$= (20 \times 14.8)$$
$$= 296\ kJ$$
$$\mathrm{\Delta}H = \mathrm{\Delta}U + \mathrm{\Delta}ngRT$$
$$= - 296 + \left( \frac{( - 3) \times 8.315 \times 298}{1000} \right)$$
$$= - 296 - 7.433$$
$$= - 303.43\ kJ$$
[IMAGE] $$\mathrm{\Delta}H_{r \times n}^{{^\circ}} = \mathrm{\Delta}H_{f}^{{^\circ}}(Hg,g) + \mathrm{\Delta}H_{r \times n}^{{^\circ}}$$
$$61.32 + \frac{1}{2}\lbrack - 303.43\rbrack$$
$$= - 90.395\ kJ$$
$\left| \mathrm{\Delta}H_{f}(HgO,s) \right| = 90.40\ kJ\ mole$.